Answer:
pH of buffer is 5.05
Explanation:
Let's assume that V mL of both solutions were added.
Number of moles of [tex]HC_{2}H_{3}O_{2}[/tex] added = [tex]\frac{0.3\times V}{1000}moles[/tex]
Number of moles of [tex]C_{2}H_{3}O_{2}^{-}[/tex] added = [tex]\frac{0.6\times V}{1000}moles[/tex]
According to Henderson-Hasselbalch equation for a buffer consisting of an weak acid ([tex]HC_{2}H_{3}O_{2}[/tex]) and it's conjugate base ([tex]C_{2}H_{3}O_{2}^{-}[/tex])-
[tex]pH=pK_{a}(HC_{2}H_{3}O_{2})+log[\frac{moles of C_{2}H_{3}O_{2}^{-}}{moles of HC_{2}H_{3}O_{2}}][/tex]
[tex]pK_{a}=-log(K_{a})[/tex]
So, [tex]pH=-log(1.8\times 10^{-5})+log[\frac{\frac{0.6\times V}{1000}}{\frac{0.3\times V}{1000}}][/tex]
So, [tex]pH=-log(1.8\times 10^{-5})+log[\frac{0.6}{0.3}][/tex]
So, pH = 5.05
